Posted by Carl on .
3. A positive helium ion He+ is released from rest at the surface of a +1000 V flat electrode. It travels in the +x-direction (in vacuum) across a 1 mm gap, passes through a small hole in a +0 V electrode that is parallel to the first one, and enters a region of magnetic field. In the region with the magnetic field, it follows a path that curves in the +y-direction. Assume the magnetic field is uniform outside the electrodes, and zero in between them. For a field strength of B = 1.0 T, what is the radius of curvature of the resulting path in meters? A helium nucleus has two protons and two neutrons
physics-can anyone solve this please -
assuming the "flat" electrode produces a uniform E field, then E=1000/.001 volts/meter
Vf^2=Vi^2+2ad=0+2(Eqd/m)d you know gap d, E, q (He is +2e, mass is 4xprotonmass)
solve for vf. That is the velocity leaving the hole.
Now in the magnetic field
Bqv=mv^2/r solve for r.